Can you explain my green underlying please.I have confused and ı dont understand why by the continuity of the orientation at the points $(0,0)$ and $(1,0)$ are also $e_{1},e_{2}$
2026-05-06 00:58:14.1778029094
The open Möbius Band is not orientable
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He simply means that the transport of basis drawn on the right of the figure is continuous. It's possible to make this more rigorous, but this is enough to make clear what the idea is.
I really don't know how to explain this more clearly than the above, but maybe rephrasing it visually is enough. Draw the basis arrows on a sheet of transparent paper. Move the paper to the right. The arrows end up as shown on the right.
More formally, I'd need your precise definitions. Technically, orientability means having a nowhere zero top form, which here amounts to a definition of the cross product of any pair of non-parallel vectors which is continuous and nonzero. (Or equivalently a notion of clockwise which is consistent.)
He's showing that, when transporting a pair of vectors smoothly around the manifold, one finds that you meet up again with a reversed notion of what you meant. But since the vectors were never parallel as we moved them around, the cross product must have been consistently pointing one way or the other if there is an orientation. This is a contradiction.